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Inductor and Capacitor Questions

Discussion in 'Electronic Basics' started by JJD, Dec 25, 2003.

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  1. JJD

    JJD Guest

    Hello all, this is my first post.

    C question:
    I understand that the equation i = C*dv/dt tells me that if there is
    an instantaneous change in voltage across a cap, that would require an
    infinite amount of current through it.
    If I charge a cap, remove it and "hold it", and put a dieletric
    material, k, between the plates, then there is an instantaneous change
    in voltage (decrease) right? q and C increase and V decreases, but i =
    0. I know that any energy lost is applied to the physical motion of
    the k material. How does all of this make sense?
    i = C*dv/dt, q = C*V, C = k*Cair(the capacitance in air)

    L question:
    If I have a simple L, Vs and switch series circuit, when I open the
    switch there should be an infinite V across the open switch, there is
    arching for a split second, but how much V is there across the switch
    in real life?

    Thanks in advance.
     
  2. John Larkin

    John Larkin Guest

    Only if you 'put' instantaneously.
    Sounds right; i = 0 by definition, since nothing is connected to the
    capacitor leads, which is where you'd measure i. What part doesn't
    make sense?

    Depends on lots of things: breakdown properties of the gap created as
    the switch opens, circuit distributed capacitance, rate of switch
    opening, amount of stored energy, stuff like that. Hundreds of volts
    peak would not be unusual with Radio Shack parts, multi kilovolts
    could be achieved with some effort. If distributed capacitance is high
    enough, it might not arc at all... it would oscillate, with amplitude
    tapering off as the energy is dissipated in various ways.

    Get some parts and try it!

    John
     
  3. The capacitance changes but the charge is unchanged (no current in or
    out), so the voltage changes.
    As much as it takes to allow the current through the inductor to decay
    in a continuous manner. Depending on how fast the contacts separate,
    the peak voltage may be in the hundreds of volts or thousands. If the
    contacts separate fast enough, the arc will extinguish before the
    current reaches zero, and the capacitance between the windings of the
    inductor will carry the inductive current, ringing with the inductance
    in a decaying sinusoid as energy dissipates in the losses in the
    inductance and capacitance and as energy radiating away from the
    resonance as radio waves.
     
  4. Ben Bradley

    Ben Bradley Guest

    There's a fun way of demonstrating this. Get an air-dielectric
    variable capacitor as was used as the tuning capacitor in old (vacuum
    tube) broadcast-band AM radios. Turn the shaft for full capacitance,
    charge it up to a couple of hundred volts or so (below the voltage
    where it arcs), disconnect the charging source, and turn the shaft to
    lower the capacitance. The charge remains the same, but the
    capacitance decreases, so the voltage increases until it arcs over.
     
  5. Mac

    Mac Guest

    I = C dv/dt applies to constant C caps.

    When you insert dielectric between the plates of a parallel plate
    capacitor, you alter C. You cannot alter q by doing this, however.

    If inserting the dielectric leads to a capacitor with more potential
    energy in it, then that means you will have to apply mechanical force to
    get the dielectric in place. And vice verse.
    The voltage will continue to rise until current flows or the backwards
    EMF is enough to shut down the current. In practice, as you seem to know,
    something always gives, in your case, the switch. So, the voltage is
    clamped by the breakdown voltage of the switch, which is probably not very
    well characterized and in any event, depends on the atmosphere around the
    switch.

    In cases where you don't want the switch to arc, you may put a
    free-wheeling diode in parallel with the inductor. The parallel diode only
    works when the current direction is always the same through the inductor.

    Diodes don't turn on instantly, so you will still get some voltage spiking
    when the switch is opened.

    Mac
     
  6. Q doesn't increase. You can tell from I = dQ/dt.
    It depends on how fast the contacts separate. During the arc, there is
    almost NO voltage across it. It could even be less that zero, because
    arcs have negative incremental resistance characteristics.
     
  7. Max_Venezia

    Max_Venezia Guest

    when you calculate something keep in mind that "zero" and "infinite"
    do not exist in the real phisical world. when time goes to zero the
    modelling change from concentrate parameters to distributed parameters
    and others parameters give a secondary effect.
     
  8. JJD

    JJD Guest

    Thank you all for the help. I am an EE student-probably you can
    tell-just starting my circuits series. I have a better understanding
    now.
     
  9. Bert Hickman

    Bert Hickman Guest

    John Woodgate wrote:
    Not quite... the negative arc characteristic only means that dV/dI is
    negative: increasing the current flow tends to decrease the voltage drop
    across the arc. However, the overall resistance of the arc will still
    always be positive.

    -- Bert --
    --
     
  10. I read in sci.electronics.design that Bert Hickman <[email protected]_
    You don't even read, or can't understand, what I wrote. What do you
    think 'incremental' means?

    READ IT!
     
  11. John Larkin

    John Larkin Guest

    Incremental negative resistance cannot produce absolute polarity
    reversal; that violates conservation of energy. The Free Energy folk
    are great fans of incremental negative slopes.

    Oscillatory ringing could (generally does) result in polarity
    reversal, but that's a whold nother issue.

    John
     
  12. I read in sci.electronics.design that John Larkin <[email protected]
    techTHISnologyPLEASE.com> wrote (in <[email protected]
    4ax.com>) about 'Inductor and Capacitor Questions', on Sat, 27 Dec 2003:
    Indeed. When I wrote 'less then zero' I was referring to the AC voltage.
    In a DC circuit, of course, the arc voltage is always greater than zero,
    but often not by much.
    Like the Gadarene Swine. Or lemmings.(;-)
    Indeed.
     
  13. John Larkin

    John Larkin Guest

    What is a less-than-zero AC voltage? My Fluke doesn't have a negative
    AC range.

    John
     
  14. I read in sci.electronics.design that John Larkin <[email protected]
    THIStechPLEASEnology.com> wrote (in <[email protected]
    4ax.com>) about 'Inductor and Capacitor Questions', on Mon, 29 Dec 2003:
    180 degrees out of phase with the supply voltage.
     
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